Fuels are burned to produce energy, and your body breaks down the food to produce nutrients for your body.
Answer is: <span>the pH value(level) is the independent variable.</span><span>
</span>
Missing question: <span>We conducted an experiment where we added 0.5 g of lactose to 5 different test tubes all containing 5 different pH levels. What is the independent variable?
In this experiment pH level changes, so results depend on different pH values.
Mass of lactose is same during experiment,so t</span>he number of molecules of product formed per minute is the <span>dependent variable.</span>
The standard enthalpy of formation for chlorine is zero but the standard entropy is larger than 0 because it is the elemental state of chlorine.
The standard enthalpy of formation for chlorine is zero because cl2 is the elemental state of chlorine and it does not require any energy for the formation of the standard state of chlorine.
The entropy of any system cannot be negative. It can only be positive or zero.
The entropy of a system will become zero only at a absolute zero temperature.
That's why the entropy of chlorine in elemental state is more than zero because absolutely zero temperature can't be obtained.
To know more about entropy, visit,
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Answer: The approximate equilibrium partial pressure of
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,

![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:

initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)


![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)

![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of
is 3.92 atm